Chapter 6, Problem 161RQ

(a)

To determine

The mass flow rate of the steam.

Explanation of Solution

Given:

The inlet pressure $\left(P_1\right)$ is $7\text{ MPa}$.

The inlet temperature $\left(T_1\right)$ is $600°\text{C}$.

The inlet velocity $\left(V_1\right)$ is $60\text{ m/s}$.

The outlet pressure $\left(P_1\right)$ is $25\text{ kPa}$.

The quality of the steam at exit $\left(x_2\right)$ is $95\%$.

The heat lost to the surroundings $\left(Q\right)$ is $20\text{ kJ/kg}$.

The area of inlet $\left(A_{inlet}\right)$ is $150{\text{ cm}}^2$

The area of outlet $\left(A_{out}\right)$ is $1400{\text{cm}}^{\text{2}}$

The outlet temperature $\left(T_1\right)$ is $600°\text{C}$.

Calculation:

At inlet:

The steam is at the state of superheated condition.

Refer Table A-6, “Superheated water”.

Obtain the inlet enthalpy $\left(h_1\right)$ and specific volume $\left(v_1\right)$ corresponding to the pressure of $7\text{MPa}$ and the temperature of $600°\text{C}$.

$\begin{array}{l}{h}_1=3650.6\text{kJ/kg}\\ v_1=0.055665{\text{m}}^3\text{/kg}\end{array}$

The turbine operates steadily. Hence, the inlet and exit mass flow rates are equal.

  ${\dot{m}}_1={\dot{m}}_2=\dot{m}$

Calculate the inlet mass flow rate.

  $\dot{m}=\frac{A_1{V}_1}{v_1}$

  $\begin{array}{c}\dot{m}=\frac{\left(150{\text{cm}}^2\right)\left(60\text{m/s}\right)}{0.055665{\text{m}}^3\text{/kg}}\\ =\frac{\left(150{\text{cm}}^2\times \frac{1\text{}{\text{m}}^2}{{10}^4{\text{cm}}^2}\right)\left(60\text{m/s}\right)}{0.055665{\text{m}}^3\text{/kg}}\\ =16.1681\text{kg/s}\\ \simeq 16.17\text{kg/s}\end{array}$

Thus, the mass flow rate of the steam is $16.17\text{kg/s}$.

(b)

To determine

The exit velocity of the steam.

Explanation of Solution

At exit:

The steam is with the quality of $95\%$.

Refer Table A-5, “Saturated water—Pressure table”.

Obtain the following corresponding to the pressure of $25\text{kPa}$.

$\begin{array}{l}{v}_f=0.001020{\text{m}}^3\text{/kg}\\ v_g=6.2034{\text{m}}^3\text{/kg}\\ h_f=271.96\text{kJ/kg}\\ h_{fg}=2345.5\text{kJ/kg}\end{array}$

Write the formula for exit enthalpy $\left(h_2\right)$ of the steam with the consideration of its quality.

  $h_2=h_f+x{h}_{fg}$

  $\begin{array}{c}{h}_2=271.96\text{kJ/kg}+95\%\left(2345.5\text{kJ/kg}\right)\\ =271.96\text{kJ/kg}+\frac{95}{100}\left(2345.5\text{kJ/kg}\right)\\ =271.96\text{kJ/kg}+2228.225\text{kJ/kg}\\ =2500.185\text{kJ/kg}\end{array}$

Write the formula for exit specific volume $\left(v_2\right)$ of the steam with the consideration of its quality.

  $v_2=v_f+x\left(v_g-v_f\right)$

  $\begin{array}{c}{v}_2=0.001020{\text{m}}^3\text{/kg}+95\%\left(6.2034{\text{m}}^3\text{/kg}-0.001020{\text{m}}^3\text{/kg}\right)\\ =0.001020{\text{m}}^3\text{/kg}+\frac{95}{100}\left(6.20238{\text{m}}^3\text{/kg}\right)\\ =0.001020{\text{m}}^3\text{/kg}+5.892261{\text{m}}^3\text{/kg}\\ =5.893281{\text{m}}^3\text{/kg}\end{array}$

The turbine operates steadily. Hence, the inlet and exit mass flow rates are equal.

  ${\dot{m}}_1={\dot{m}}_2=\dot{m}$

Write the formula for exit mass flow rate.

  $\dot{m}=\frac{A_2{V}_2}{v_2}$

Rearrange the above Equation to obtain exit velocity $\left(V_2\right)$.

  $V_2=\frac{\dot{m}{v}_2}{A_2}$

  $\begin{array}{c}{V}_2=\frac{\left(16.17\text{kg/s}\right)\left(5.893281{\text{m}}^3\text{/kg}\right)}{1400{\text{cm}}^2}\\ =\frac{\left(16.17\text{kg/s}\right)\left(5.893281{\text{m}}^3\text{/kg}\right)}{1400{\text{cm}}^2\times \frac{1\text{}{\text{m}}^2}{{10}^4{\text{cm}}^2}}\\ =680.6739\text{m/s}\end{array}$

Thus, the exit velocity of the steam is $680.6739\text{m/s}$.

(c)

To determine

The power output of the turbine.

Answer to Problem 161RQ

The power output of the turbine is $14562\text{kW}$.

Explanation of Solution

Consider the steam flows at steady state. Hence, the inlet and exit mass flow rates are equal.

  ${\dot{m}}_1={\dot{m}}_2=\dot{m}$

Write the energy rate balance equation for one inlet and one outlet system.

  $\left[{\dot{Q}}_{\text{1}}+{\dot{W}}_{\text{1}}+\dot{m}\left(h_1+\frac{V_1{}^2}{2}+g{z}_1\right)\right]-\left[{\dot{Q}}_{\text{2}}+{\dot{W}}_{\text{2}}+\dot{m}\left(h_2+\frac{V_2{}^2}{2}+g{z}_2\right)\right]=\Delta {\dot{E}}_{\text{system}}$        (VI)

Here, the rate of heat transfer is $\dot{Q}$, the rate of work transfer is $\dot{W}$, the enthalpy is $h$ and the velocity is $V$, the gravitational acceleration is $g$, the elevation from the datum is $z$ and the rate of change in net energy of the system is $\Delta {\dot{E}}_{\text{system}}$; the suffixes 1 and 2 indicates the inlet and outlet of the system.

The refrigerant flows at steady state through the compressor. Hence, the rate of change in net energy of the system becomes zero.

  $\Delta {\dot{E}}_{\text{system}}=0$

Heat loss occurs at the rate of $20\text{kJ/kg}$. Neglect the potential energy changes. Here, the work done is by the system (turbine) and the work done on the system is zero i.e. ${\dot{W}}_1=0$.

The Equations (VI) reduced as follows to obtain the work output $\left({\dot{W}}_2\right)$.

  $\begin{array}{l}\left[0+0+\dot{m}\left(h_1+\frac{V_1{}^2}{2}+0\right)\right]-\left[{\dot{Q}}_2+{\dot{W}}_2+\dot{m}\left(h_2+\frac{V_2{}^2}{2}+0\right)\right]=0\\ \dot{m}\left(h_1+\frac{V_1{}^2}{2}\right)-\left[{\dot{Q}}_2+{\dot{W}}_2+\dot{m}\left(h_2+\frac{V_2{}^2}{2}\right)\right]=0\\ \dot{m}\left(h_1+\frac{V_1{}^2}{2}\right)={\dot{Q}}_2+{\dot{W}}_2+\dot{m}\left(h_2+\frac{V_2{}^2}{2}\right)\\ {\dot{W}}_2=\dot{m}\left(h_1+\frac{V_1{}^2}{2}\right)-\dot{m}\left(h_2+\frac{V_2{}^2}{2}\right)-{\dot{Q}}_2\end{array}$

  $\begin{array}{l}{\dot{W}}_2=\dot{m}\left(h_1-h_2+\frac{V_1{}^2-V_2{}^2}{2}\right)-{\dot{Q}}_2\\ {\dot{W}}_2=-\dot{m}\left(h_2-h_1+\frac{V_2{}^2-V_1{}^2}{2}\right)-{\dot{Q}}_2\end{array}$        (VII)

Here, ${\dot{Q}}_2=\dot{m}{Q}_2$.

Rewrite the Equation (VII) as follows.

  ${\dot{W}}_2=-\dot{m}\left(h_2-h_1+\frac{V_2{}^2-V_1{}^2}{2}\right)-\dot{m}{Q}_2$        (VIII)

  $\begin{array}{c}{\dot{W}}_2=-16.17\text{kg/s}\left[\begin{array}{l}2500.185\text{kJ/kg}-3650.6\text{kJ/kg}\\ +\frac{{\left(680.6739\text{m/s}\right)}^2-{\left(60\text{m/s}\right)}^2}{2}\end{array}\right]-16.17\text{kg/s}\left(20\text{kJ/kg}\right)\\ =\left[\begin{array}{c}-16.17\text{kg/s}\left(-1150.415\text{kJ/kg}+229858.4791{\text{m}}^2{\text{/s}}^2\times \frac{1\text{}\text{kJ/kg}}{1000{\text{m}}^2{\text{/s}}^2}\right)\\ -323.4\text{kJ/s}\end{array}\right]\\ =-16.17\text{kg/s}\left(-920.5565\text{kJ/kg}\right)-323.4\text{kJ/s}\\ =\text{14885}\text{.3989}\text{kJ/s}-323.4\text{kJ/s}\end{array}$

  $\begin{array}{c}=14561.9989\text{kJ/s}\times \frac{1\text{kW}}{1\text{kJ/s}}\\ \simeq 14562\text{kW}\end{array}$

Thus, the power output of the turbine is $14562\text{kW}$.